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Ch. 11—Intermolecular Forces, Liquids, and Solids Here are some basic properties of the 3 phases of matter…

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Presentation on theme: "Ch. 11—Intermolecular Forces, Liquids, and Solids Here are some basic properties of the 3 phases of matter…"— Presentation transcript:

1 Ch. 11—Intermolecular Forces, Liquids, and Solids Here are some basic properties of the 3 phases of matter…

2 Intermolecular Forces The forces holding solids and liquids together are called intermolecular forces. –Intermolecular forces are much weaker than ionic or covalent bonds. Example: 16 kJ/mol to vaporize HCl compared to 431 kJ/mol to break HCl into its elements. –When a substance melts or boils, the intermolecular forces are broken (not the covalent bonds).

3 Intermolecular Forces Boiling point reflects intermolecular force strength. - A high boiling point indicates strong attractive forces. - A high melting point also reflects strong attractive forces. There are 4 basic types of intermolecular forces… 1) ion-dipole forces 2) dipole-dipole forces 3) London dispersion forces 4) hydrogen bonding…(a special type of dipole force)

4 Ion-Dipole Forces Interaction between an ion and a dipole. Strongest of all intermolecular forces. Example: Na + and Cl - ions dissolved in water. Dipole-Dipole Forces Exist between neutral polar molecules. Weaker than ion-dipole forces. If two molecules have about the same mass and size, then dipole-dipole forces increase with increasing polarity.

5 Dipole-Dipole Forces London Dispersion Forces Weakest of all intermolecular forces… London dispersion forces exist between all molecules! One molecule’s electron cloud can become distorted causing an instantaneous dipole & that can induce another instantaneous dipole in an adjacent molecule (or atom). The forces between instantaneous dipoles are called London dispersion forces.

6 London Dispersion Forces “instantaneous dipoles” The larger the molecule (the greater the number of electrons) the more polarizable or the easier it is to create instantaneous dipoles. London dispersion forces increase as molecular weight increases.

7 London Dispersion Forces London dispersion forces depend on the shape of the molecule. The greater the surface area available for contact, the greater the dispersion forces. London dispersion forces between spherical nonpolar molecules are lower than the forces between long nonpolar molecules. less surface area, less force

8 Hydrogen Bonding A special case of dipole-dipole forces. By experiments, the boiling pts. of compounds with H-F, H-O, and H-N bonds are abnormally high. The intermolecular forces are therefore abnormally strong. H-bonding requires… 1) H bonded to a small, highly electronegative element (most important for compounds of F, O, and N) 2) an unshared electron pair on a nearby small highly electronegative ion or atom (usually F, O, or N on another molecule).

9 Hydrogen Bonding Examples 1) 2) 3) abnormally high B.P. polar molecules nonpolar molecules

10 Properties of Liquids Liquids vary in viscosity. - Viscosity is the resistance of a liquid to flow. -The “thicker” the liquid, the more viscosity. -The stronger the intermolecular forces, the higher the viscosity. -Typically, as temperature increases, viscosity decreases.

11 Properties of Liquids Surface Tension -Acts as a thin skin. -Bulk molecules (those in the liquid) are equally attracted to their neighbors. Surface molecules are only attracted inwards towards the bulk molecules. Therefore, surface molecules are packed more closely than bulk molecules. -Surface tension is the amount of energy required to increase the surface area of a liquid. -The stronger the intermolecular forces, the greater the surface tension.

12 Phase Changes & Energy Endothermic: melting, evaporating/boiling & sublimation Exothermic: freezing, condensation, & deposition

13 Phase Changes & Energy heat of vaporization: the heat energy required to evaporate a given mass of liquid at a constant temperature heat of fusion: the heat energy required to melt a given mass of solid at a constant temperature Heating Curve The temperature, (average KE), during a phase change (such as boiling) does not change! Any heat added during boiling gives more molecules enough energy to escape the liquid.

14 Phase Changes & Energy Generally, it will take more heat to vaporize a liquid than to melt a solid… (∆H (vap) > ∆H (fusion) ) Why? - Every intermolecular bond is broken when vaporizing, but only some of the intermolecular forces break when melting solids.

15 Liquefying Gases Gases can be liquefied by increasing pressure at some temperature. - Critical temperature: the highest temperature at which a substance can remain a liquid regardless of the pressure applied. - Critical pressure: the pressure needed at the critical temperature. Notice: As intermolecular attractions increase, critical temp, & pressure increase.

16 Vapor Pressure and Boiling Liquids Some of the molecules on the surface of a liquid have enough energy to escape the attraction of the liquid. These molecules move into the gas phase. As the number of molecules in the gas phase increases, some of the gas phase molecules strike the surface and return to the liquid. After some time the pressure of the gas will be constant at the vapor pressure. Dynamic Equilibrium: the point when as many molecules escape the surface as strike the surface. Vapor pressure is the pressure exerted when the liquid and vapor are in dynamic equilibrium.

17 Vapor Pressure and Boiling Liquids Liquids boil when the external pressure equals the vapor pressure. Temperature of the boiling point increases as pressure increases. Normal boiling point is the boiling point at 760 mmHg (1 atm). A substance with a high vapor pressure is said to be volatile. It readily evaporates.

18 Vapor Pressure and Boiling Liquids Two ways to get a liquid to boil: 1) increase temperature 2) decrease pressure (vacuum pump) Pressure cookers operate at high pressure. - At high pressure the boiling point of water is higher than at 1 atm. - Therefore, there is a higher temperature at which the food is cooked, reducing the cooking time required. - Boiling points increase with molecular weight as long (as the intermolecular forces are similar.) - Example: CH 4 (m.w. = 16) < C 2 H 6 (m.w. = 30) < C 3 H 8 (m.w. = 44) [ lowest B.P.] [highest B.P.]

19 Phase Diagrams Shows the relationship between the 3 phases of matter at various temperatures and pressures. Triple Point: All 3 phases of matter at equilibrium. Critical Point: The highest temperature at which the liquid phase can exist.

20 Phase Diagrams of H 2 O and CO 2 Notice the slope of the solid–liquid equilibrium line. This indicates that water expands when it freezes and CO 2 contracts when it freezes.

21 Characteristics of Solids There are 2 general classifications of solids: -Amorphous: No pattern to the arrangement of particles. Their melting point is over a wide range of temperatures. They just get softer and softer when heated. (Examples- glass, plastic, wax) - Crystalline: Well-ordered, definite arrangement of atoms. Crystals have a repeated structure and a melting point at a very narrow range of temperatures. (Examples- metals, H 2 O, diamond)

22 Crystalline Solids There are four types of crystalline solid: - Molecular (formed from molecules) - usually soft with low melting points and poor conductivity. - Covalent network (formed from atoms) - very hard with very high melting points and poor conductivity. - Ionic(formed form ions) - hard, brittle, high melting points and poor conductivity. - Metallic (formed from metal atoms) - soft or hard, high melting points, good conductivity, malleable and ductile.

23 Bonding in Crystalline Solids Metallic bonds are formed from metal nuclei floating in a sea of electrons.

24 Crystalline Solids Covalent Network IonicMetallic Molecular

25 Structure of Solids Unit Cell: The smallest repeating unit in a crystal is a unit cell. - A unit cell is the smallest unit with all the symmetry of the entire crystal. Three-dimensional stacking of unit cells is the crystal lattice.

26 Unit Cell There are three common types of unit cell. –Primitive (simple) cubic: atoms at the corners of a simple cube, each atom shared by 8 unit cells –Body-centered cubic (bcc): atoms at the corners of a cube plus one in the center of the body of the cube, corner atoms shared by 8 unit cells, center atom completely enclosed in one unit cell –Face-centered cubic (fcc): atoms at the corners of a cube plus one atom in the center of each face of the cube, corner atoms shared by 8 unit cells, face atoms shared by 2 unit cells

27 Unit Cell

28 # of Atoms/Unit Cell

29 simple cubic unit cell = 1 atom body centered cubic = 2 atoms face-centered cubic = 4 atoms

30 Close Packing of Spheres We rationalize maximum intermolecular force in a crystal by the close packing of spheres. A crystal is built up by placing close packed layers of spheres on top of each other. - Molecules can be modeled by spheres. - Atoms and ions are spheres. When spheres are packed as closely as possible, there are small spaces between adjacent spheres. The spaces are called interstitial holes. hole

31 Close Packing of Spheres There is only one place for the second layer of spheres…in the hole. There are two choices for the third layer of spheres: - Third layer eclipses the first (ABAB arrangement). This is called hexagonal close packing (hcp). - Third layer is in a different position relative to the first (ABCABC arrangement). This is cubic close packing (ccp).

32 Coordination Number In both types of closest packing, hcp & ccp, each sphere is surrounded by 12 other spheres (6 in one plane, 3 above and 3 below). Coordination number: the number of spheres directly surrounding a central sphere. By comparison…- Simple (primitive) cubic structure= 6 - Body-centered cubic structure= 8 - Face-centered cubic structure= 12 Hexagonal and cubic close packing are different from the cubic unit cells. If unequally sized spheres are used, the smaller spheres are placed in the “interstitial holes”.

33 Interstitial Holes In the face centered cubic (fcc) cell there is more than one type of “hole”. If the octahedral holes are filled, the structure above results, with a 1:1 count for the two types of ions in the salt. If the tetrahedral holes are filled, a different structure exits. It will have a 2:1 count for the ions in the salt. In the figure below, The left shows the structure of NaCl and the right that of CaF 2. (By the way, not all of the “holes” need to be filled up. Ion size determines this.)

34 Calculations Using Unit Cells There are 2 basic calculations involving unit cells: 1) Finding the density of an element given the type of unit cell and the length of one side of the lattice. Density = mass/volume mass = volume = LxWxH = (length of one side) 3 … we will use “a” as the length of one side, so… V= a 3 Putting them together…Density= (# of atoms in the unit cell) x (1 mole)_____ (6.02 x 10 23 atoms) x (formula mass) (1 mole) (# of atoms in the unit cell) x _(formula mass)_ (6.02 x 10 23 atoms)(a 3 ) Remember: simple cubic = 1 atom; body centered = 2 atoms; face-centered = 4 atoms

35 Calculations Using Unit Cells 2) Finding the radius of the atom given the type of unit cell and the length of one side of the lattice. (This can be a little more complicated to derive a nice formula.) By looking at the unit cells, we can determine how the length of one side, (a), is related to the radius, (r), of one atom… Simple Cubic: a = 2r or r = ½a …easy to see! Face Centered: a 2 +a 2 = (4r) 2 … Pythagorean’s Theorem Simplifying… 2a 2 =16r 2 … a 2 =8r 2 … a = r (√8) Body Centered: We need a better 3-D view in order to derive a formula! a a a

36 Calculations Using Unit Cells A body-centered lattice is slightly trickier than the face-centered lattice because our diagonal doesn’t lie on the face of the cube. Instead, it lies within the body of the cube. We will also assume that the particles come in contact with each other unlike this drawing. a Solving for the triangle in blue… c 2 = a 2 + b 2 … ( 4r) 2 = a 2 + b 2 We don’t have a value for “b”, but we can recognize that it is also a hypotenuse of a right triangle… b 2 = a 2 +a 2 Substituting… (4r) 2 = a 2 + a 2 + a 2 Simplifying…16r 2 = 3a 2 Solving for “a”… a = r(√5⅓ ) a a Now we get to do practice problems! (body-centered lattice)


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